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integral of sqrt(a-x^2)

解答

\int a - x^{2} d x

解答

\frac{1}{2} a (arcsin (\frac{1}{a} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{a} x))) + C

求解步骤

\int a - x^{2} d x

a - x^{2} = a (1 - \frac{x ^{2}}{a})

= \int a (1 - \frac{x ^{2}}{a}) d x

使用指数法则:

\frac{a ^{n}}{b ^{n}} = (\frac{a}{b})^{n}

\frac{x ^{2}}{a} = (\frac{x}{a})^{2}

= \int a (1 - (\frac{x}{a})^{2}) d x

(\frac{x}{a})^{2} = \frac{x ^{2}}{a}

= \int a (1 - \frac{x ^{2}}{a}) d x

使用指数法则:

(a \cdot b)^{n} = a^{n} \cdot b^{n}

a (1 - \frac{x ^{2}}{a}) = a (1 - \frac{x ^{2}}{a})

= \int a 1 - \frac{x ^{2}}{a} d x

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= a \cdot \int 1 - \frac{x ^{2}}{a} d x

使用三角换元法

= a \cdot \int a cos^{2} (u) d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= a a \cdot \int cos^{2} (u) d u

使用三角恒等式改写

= a a \cdot \int \frac{1 + cos ( 2 u )}{2} d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= a a \frac{1}{2} \cdot \int 1 + cos (2 u) d u

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= a a \frac{1}{2} (\int 1 d u + \int cos (2 u) d u)

\int 1 d u = u

\int cos (2 u) d u = \frac{1}{2} sin (2 u)

= a a \frac{1}{2} (u + \frac{1}{2} sin (2 u))

u = arcsin (\frac{1}{a} x)

代回

= a a \frac{1}{2} (arcsin (\frac{1}{a} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{a} x)))

使用根式运算法则:

a a = a

a a = a

= \frac{1}{2} a (arcsin (\frac{1}{a} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{a} x)))

解答补常数

= \frac{1}{2} a (arcsin (\frac{1}{a} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{a} x))) + C

作图

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